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How would you evaluate the following indefinite integral? In fact, I did evaluate $\int \frac{\cos{t}}{t} dt$ by parametric integration and then I thought of this variant. $$\int \frac{t}{\cos{t}} dt$$

Here you may find the result given by W|A.

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Did you evaluated the primitive of $\cos t/t$? – enzotib Aug 13 '12 at 14:43
@enzotib, that's actually easier, since it's just the cosine integral. This one will require the inverse tangent integral. – J. M. Aug 13 '12 at 15:12
@J.M.: yes, but by definition of cosine integral, not actually evaluating something, like the OP claims. – enzotib Aug 13 '12 at 15:15
@enzotib, maybe OP went further and expressed it in terms of the incomplete gamma function... :D – J. M. Aug 13 '12 at 15:20
@enzotib: In fact, I did try it. – OFFSHARING Aug 14 '12 at 7:18

2 Answers 2

up vote 4 down vote accepted

Let's start with $\ \displaystyle f(x):=\log(\tan(x/2))\ $ then : $$f'(x)=\frac{\tan(x/2)^2+1}{2\tan(x/2)}=\frac{\sin(x/2)^2+\cos(x/2)^2}{2\cos(x/2)^2\tan(x/2)}=\frac 1{\sin(x)}$$

so that $\ f'\left(t+\frac {\pi}2\right)=\dfrac 1{\cos(t)}$.

We want (ignoring integration constants up to the end) : $$\int \dfrac t{\cos(t)}\,dt=\int tf'\left(t+\frac {\pi}2\right) dt=\left[tf\left(t+\frac {\pi}2\right)\right]-\int f\left(t+\frac {\pi}2\right)\,dt$$ Setting $\ u:=\tan\bigl(\frac x2\bigr)\ $ so that $\ dx=\dfrac {2\;du}{1+u^2}$ we rewrite the integral of $f$ as : $$\int f(x)\;dx=\int \log\left(\tan\left(\frac x2\right)\right)\;dx=2\int \frac{\log(u)}{1+u^2}\;du$$ $$=\left[2\log(u)\arctan(u)\right]-2\int \frac {\arctan(u)}u\,du=2\left[\log(u)\arctan(u)-\rm{Ti}_2(u)\right]$$ with $\rm{Ti}_2$ the inverse tangent integral proposed by J.M. (see Lewin 1981 "Polylogarithms and associated functions" ch. 2 for more information). $\rm{Ti}_2$ may be rewritten as Clausen functions or as complex polylogarithms.

We want $\ \int f\bigl(t+\frac {\pi}2\bigr)\,dt\ $ so that $\ u:=\tan\bigl(\frac t2+\frac {\pi}4\bigr)\ $ and $$\int f\left(t+\frac {\pi}2\right)\,dt=2\left[\log\left(\tan\left(\frac t2+\frac {\pi}4\right)\right)\left(\frac t2+\frac {\pi}4\right)-\rm{Ti}_2\left(\tan\left(\frac t2+\frac {\pi}4\right)\right)\right]$$

getting : $$\int \dfrac t{\cos(t)}\,dt=t\left[\log\left(\tan\left(\frac t2+\frac {\pi}4\right)\right)\right]-2\left[\log\left(\tan\left(\frac t2+\frac {\pi}4\right)\right)\left(\frac t2+\frac {\pi}4\right)-\rm{Ti}_2\left(\tan\left(\frac t2+\frac {\pi}4\right)\right)\right]$$

and finally : $$\int_0^t \dfrac x{\cos(x)}\,dx=-\frac {\pi}2\log\left(\tan\left(\frac t2+\frac {\pi}4\right)\right)+2\rm{Ti}_2\left(\tan\left(\frac t2+\frac {\pi}4\right)\right)+C$$ where the additional constant $\ C=-2\;\rm{Ti}_2(1)=-2K\quad$ ($K$ is the Catalan constant).

To rewrite this with Clausen functions we may use (4.31) of Lewin's reference : $$\rm{Ti}_2(\tan \theta)=\theta\log(\tan\theta)+\frac 12\left(\rm{Cl}_2(2\theta)-\rm{Cl}_2(\pi-2\theta)\right)$$

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Thanks for following through, again. :) I had forgotten about the Clausen function... good thing you mentioned it too, for the benefit of those who want to avoid using complex numbers in real results. – J. M. Aug 14 '12 at 5:19
@Raymond Manzoni: Clausen functions? I didn't know they have a name. :-) Thanks. – OFFSHARING Aug 14 '12 at 7:13
@Chris'sister: Glad you liked my 'all real' solution ! Clausen functions are indeed interesting since you may write them as $\displaystyle \rm{Cl}_{2n}(\theta)=\sum_{n=1}^\infty \frac {\sin(n\theta)}{n^{2n}}$ and $\displaystyle \rm{Cl}_{2n+1}(\theta)=\sum_{n=1}^\infty \frac {\cos(n\theta)}{n^{2n+1}}$as found at Mathworld (I think that wikipedia's 'General definition' should rather be something like ... – Raymond Manzoni Aug 14 '12 at 8:51
$\displaystyle \rm{Cl}_s(\theta)=\sum_{n=1}^\infty \frac {\sin(n\theta+s\pi/2)}{n^s}$ if we accept to change some signs !). These series are frequent in Fourier series and the other half of the cases is handled by the Bernoulli polynomials. – Raymond Manzoni Aug 14 '12 at 8:52
@Raymond Manzoni: I usually like the solutions from which I have something to learn as this solution is. – OFFSHARING Aug 14 '12 at 8:55

Note that $$ \frac{1}{\cos t}= \frac{2}{e^{it}+e^{-it}}= \frac{2e^{it}}{1+e^{2it}} $$ so integration by parts gives $$ \int\frac{t}{\cos t}dt= \int 2t\frac{e^{it}}{1+e^{2it}}dt= \int 2t\frac{(-i)d(e^{it})}{1+e^{2it}}= \int -2itd(\arctan(e^{it}))= $$ $$ -2it\arctan(e^{it})-\int\arctan(e^{it})d(-2it)= -2it\arctan(e^{it})+2i\int\arctan(e^{it})dt $$ It is remains to compute the last integral $$ \int\arctan(e^{it})dt= \int\sum\limits_{k=1}^\infty \frac{(-1)^k(e^{it})^{2k+1}}{2k+1}dt= \int\sum\limits_{k=1}^\infty \frac{(ie^{it})^{2k+1}}{i(2k+1)}dt= \int\frac{1}{2i}\left(\sum\limits_{k=1}^\infty\frac{(ie^{it})^k}{k}- \sum\limits_{k=1}^\infty\frac{(-ie^{it})^k}{k}\right)dt= \frac{1}{2i}\sum\limits_{k=1}^\infty\int\frac{(ie^{it})^k}{k}dt- \frac{1}{2i}\sum\limits_{k=1}^\infty\int\frac{(-ie^{it})^k}{k}dt= $$ $$ \frac{1}{2i}\sum\limits_{k=1}^\infty\frac{(ie^{it})^k}{ik^2}- \frac{1}{2i}\sum\limits_{k=1}^\infty\frac{(-ie^{it})^k}{ik^2}= \frac{1}{2}\left(\mathrm{Li}_2(-ie^{it})-\mathrm{Li}_2(ie^{it})\right) $$ The final result is $$ \int\frac{t}{\cos t}=-2it\arctan(e^{it})+i(\mathrm{Li}_2(-ie^{it})-\mathrm{Li}_2(ie^{it})) $$

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As I've already noted, one might choose to use the inverse tangent integral instead of the dilogarithm in this case... – J. M. Aug 13 '12 at 16:07
I saw your comment after I finished writing this answer – Norbert Aug 13 '12 at 16:34
@Norbert: thanks for your answer. – OFFSHARING Aug 14 '12 at 7:12
I like all your provided answers. (+1) – OFFSHARING Aug 14 '12 at 9:17
Thanks @Chris'sister, glad to see nice questions – Norbert Aug 14 '12 at 9:45

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